testing collinearity with algorithm collinear7

조회 수: 11(최근 30일)
Jürgen
Jürgen 2012년 12월 8일
Hi,
I was looking for an algorithm to test collinearity and found : http://blogs.mathworks.com/loren/2008/06/06/collinearity/
function tf = collinear7(p1,p2,p3)
mat = [p1(1)-p3(1) p1(2)-p3(2); ...
p2(1)-p3(1) p2(2)-p3(2)];
tf = det(mat) == 0;
works fine if you test it with eg: p1 = [1 1]; p2 = [3.5 3.5]; p3 = [-7.2 -7.2]; collinear(p1,p2,p3) ans=1
for testing I choose: p1 = [ 123.83 -205.59] p2 = [ 138.38 -990.38] p3=(p1+p2)/2 p3 = 131.11 -597.98
collinear(p1,p2,p3) ans = 0 here I was suprised because two points and the middle point should be collinear no? so I tried it again with p1=[1.5 -5.3] p2=[ 5 -3.8] p3=(p1+p2)/2 collinear(p1,p2,p3) ans = 1 so now it is ok again???
so I tested another approach slightky changed the code from the site
function tf = collinear1(p1,p2,p3)
m = slope(p1,p2);
b = intercept(p1,p2);
y = (m*p3(1)+b) ;
tf=abs(y- p3(2))<0.000001;
end
function m = slope(p1,p2)
m = (p2(2)-p1(2))/(p2(1)-p1(1));
end
function b = intercept(p1,p2)
m = slope(p1,p2);
b = p1(2)-m*p1(1);
end
with p1 = [ 123.83 -205.59] p2 = [ 138.38 -990.38] p3=(p1+p2)/2
collinear1(p1,p2,p3) is 1=> so collinear?
I can only think of numerical errors to explain this or am I doing something completely wrong?
regards,J

채택된 답변

Matt J
Matt J 2012년 12월 8일
편집: Matt J 2012년 12월 12일
Yes, it's a numerical problem. The first version collinear7 uses the criterion
det(mat) == 0
which demands both perfect collinearity among the points and a perfect determinant calculation, with no tolerance for floating point errors. The other versions apply a tolerance on floating point inaccuracies.

추가 답변(0개)

범주

Find more on Creating and Concatenating Matrices in Help Center and File Exchange

태그

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by